Question

prove that the difference of square of odd integers (numbers) are divisible by 8

Answer 1

Hi Dear !______________________________________________________________Let two odd numbers are 2x+1 , 2y+1
Where x and y are integers
The difference of their  squares is :
(2x+1)²      -      (2y+1)²
(4x²+1+4x)-(4y²+1+4y)
4x²+1+4x-4y²-1-4y
4x²+4x-4y²-4y
4[x²-x-y²-y]
4[x(x+1)-y(y+1)]
We know that x and x+1 are consecutive numbers 
and one of em is divisible by 2 
so we can write x(x+1) is divisible by two 
i.e. let      x(x+1)= 2p
similarly ,y(y+1)= 2q
difference = 4 [2p-2q]
difference = 8 [p-q]
Hence it is divisible by 8 
____________________________________________________________

Answer 2

Hey frnd here is the simple method

Let x and x+2 are two odd integer

(x+2)²-x² is divisible by 8
Step 1.
Let us prove that result is true for x= 1
(3)²-(1)² =9-1= 8
8 is divisible by 8
Result is true for x= 1

Step 2.
Let us assume that result is true for
x=k
(k+2)²- (k)²= 8m
(k+2+k)(k+2-k) =8m
(2k+2)×2=8m
(k+1)= 2m -----(1)
for some constant m

Step 3.
We have to prove that result is true for
(k+2)

(k+4)²- (k+2)²
k²+16+8k-k²-4-4k
→12+4k
→4(k+3)
→4(k+1+2)
→4(2m+2)
→8(m+1)
8(m+1) is divisible by 8
Hence proved.